Consider the given data : (a) HCl(g) + 10 H2O(l) -> HCl.10 H2O ΔH = -69.01 kJ mol-1 (b) HCl(g) + 40 H2O(l) -> HCl.40 H2O ΔH = -72.79 kJ mol-1 Choose the correct statement :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q52 – JEE-MAIN 2025

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Home Chemistry Chemical ThermodynamicsQuestion #308

Consider the given data : \ (a)

H C l (g) + 10 H_{2} O (l) \to H C l .10 H_{2} O Δ H = - 69.01 {kJ mol}^{- 1}

\ (b)

H C l (g) + 40 H_{2} O (l) \to H C l .40 H_{2} O Δ H = - 72.79 {kJ mol}^{- 1}

\ Choose the correct statement :

Clue 1 of 3: Enthalpy of Solution

The enthalpy of solution is the heat change when one mole of a substance dissolves in a specified amount of solvent.

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Ninja Strategy•Direct Data Interpretation & Definition Check

Directly compare the given enthalpy values to assess the dependence on solvent amount, calculate dilution enthalpy using Hess's Law, and recall definitions of exothermic/endothermic and heat of formation.

🧠 Full Solution Path

Step 1: Analyze the given reactions and definitions✦ Active

The given reactions (a) and (b) represent the enthalpy of solution of $H C l$ gas in different amounts of water:

(a) H C l (g) + 10 H_{2} O (l) \to H C l .10 H_{2} O Δ H_{a} = - 69.01 {kJ mol}^{- 1}

(b) H C l (g) + 40 H_{2} O (l) \to H C l .40 H_{2} O Δ H_{b} = - 72.79 {kJ mol}^{- 1}

We need to evaluate the given statements regarding heat of solution, heat of dilution, and the nature of the dissolution process.

Step 2: Evaluate each statement○ Expand

1. **Statement 1: The heat of solution depends on the amount of solvent.** Comparing $Δ H_{a}$ and $Δ H_{b}$ , the enthalpy of solution changes from $- 69.01 {kJ mol}^{- 1}$ to $- 72.79 {kJ mol}^{- 1}$ when the amount of solvent increases. This statement is **correct**.

2. **Statement 2: The heat of dilution for the $H C l (H C l .10 H_{2} O to H C l .40 H_{2} O)$ is $3.78 {kJ mol}^{- 1}$ .** The dilution process is $H C l .10 H_{2} O + 30 H_{2} O (l) \to H C l .40 H_{2} O$ . Using Hess's Law, $Δ H_{d i l u t i o n} = Δ H_{b} - Δ H_{a} = (- 72.79) - (- 69.01) = - 3.78 {kJ mol}^{- 1}$ . The statement gives a positive value, so this statement is **incorrect**.

3. **Statement 3: Dissolution of gas in water is an endothermic process.** Both $Δ H_{a}$ and $Δ H_{b}$ are negative, indicating that the dissolution of $H C l$ gas in water is an exothermic process (releases heat). This statement is **incorrect**.

4. **Statement 4: The heat of formation of $H C l$ solution is represented by both (a) and (b).** The heat of formation is the enthalpy change when a compound is formed from its constituent elements in their standard states. Reactions (a) and (b) start with $H C l$ gas, not its elements ( $H_{2}$ and $C l_{2}$ ). Therefore, these reactions represent the heat of solution, not the heat of formation. This statement is **incorrect**.

💡 Teacher's Secret Hint

Remember that the heat of formation specifically refers to formation from elements in their standard states.

Step 3: Conclude the correct statement○ Expand

Based on the evaluation, only Statement 1 is correct.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 52 – JEE-MAIN 2025

Clue 1 of 3: Enthalpy of Solution

🧠 Full Solution Path